Question

cho tam giác ABC có AB=4, BC=7, AC=9. tính cos(B+C)

Answer 1

\(cosB=\frac{AB^2+BC^2-AC^2}{2AB.BC}=-\frac{2}{7}\)

\(\Rightarrow sinB=\sqrt{1-cos^2B}=\frac{3\sqrt{5}}{7}\)

\(cosC=\frac{AC^2+BC^2-AB^2}{2AC.BC}=\frac{19}{21}\)

\(\Rightarrow sinC=\sqrt{1-cos^2C}=\frac{4\sqrt{5}}{21}\)

\(cos\left(B+C\right)=cosB.cosC-sinB.sinC=...\)

Answer 2

Nguyễn Việt Lâm giúp mk vs.